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					                                           IIT-JEE 2011 EXAMINATION                                                       PAPER - 1
                                                       (HELD ON : 10 - 04 - 2011)

                                                    PART - I (CHEMISTRY)
                                           SECTION - I (TOTAL MARKS : 21)
                                   (Single Correct Answer Type)
     This section contains 7 multiple choice questions. Each question has four choices (A), (B), (C) and
     (D), out of which ONLY ONE is correct.
1.   Geometrical shapes of the complexes formed by the reaction of Ni2+ with Cl–, CN– and H2O respectively,
     are -
     (A) octahedral, tetrahedral and square planar
     (B) tetrahedral, square planar and octahedral
     (C) square planar, tetrahedral and octahedral
     (D) octahedral, square planar and octahedral
1.   Ans.(B)



              EN
     [NiCl 4 ]2- , [Ni(CN)4 ]2- , [Ni(H 2O) 6 ]2+




      Cl
             Cl

             Ni
                           2–


                                    NC                 CN
                                                             2–
                                                                      H2 O

                                                                      H2 O
                                                                                 OH2

                                                                                Ni
                                                                                     OH2
                                                                                         OH2
                                                                                                 2+


                      Cl                      Ni
            LL
            Cl                      NC                 CN                       OH2
            sp3                            dsp     2
                                                                               sp3d2
         tetrahedral                  square planar                        octahedral

2.   AgNO3 (aq.) was added to an aqueous KCl solution gradually and the conductivity of the solution
     was measured. the plot of conductance (L) versus the volume of AgNO3 is -
           A


     L                                 L                                 L                                L


             volume                              volume                             volume                             volume
              (P)                                 (Q)                                (R)                                (S)

     (A) (P)                          (B) (Q)                           (C) (R)                            (D) (S)
2.   Ans.(D)
     As AgNO3 is added
     K+ + Cl– + Ag+ + NO3 ¾® AgCl(s)¯ + K+ + NO3
                        –                      –


     Number of ions are approximately constant, when AgNO3 is added till equivalence point. After
     equivalence point number of ions, increases and hence conductivity.




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3.   Bombardment of aluminium by a-particle leads to its artificial disintegration in two ways, (i) and (ii)
     as shown. Products X, Y and Z respectively are :

                                       27             (ii)         30
                                       13
                                          Al                       15
                                                                      P   + Y
                                        (i)

                                   30                           30
                                   14 Si       + X              14Si      + Z

     (A) proton, neutron, positron                                     (B) neutron, positron, proton
     (C) proton, positron, neutron                                     (D) positron, proton, neutron
3.   Ans.(A)
     (i)    27
            13   Al + 2 He ¾¾ 14 Si + 1P
                      4
                            ® 30      1



     (ii)   27
            13


               ENAl + 2 He ¾¾ 15 P + 1 n
                      4
                            ® 30


                                  30
                                  14
                                     0




                                       Si + 1 e
     X ® proton, Y ® neutron, Z ® positron
                                            0
             LL
4.   Extra pure N2 can be obtained by heating
     (A) NH3 with CuO                                                  (B) NH4NO3
     (C) (NH4)2Cr2O7                                                   (D) Ba(N3)2
4.   Ans.(D)
            A

     Extra pure N2 can be obtained by heating of Ba(N3)2
              D
                ®
     Ba(N3)2 ¾¾ Ba+ 3N2­


5.   Among the following compounds, the most acidic is
     (A) p-nitrophenol                                                 (B) p-hydroxybenzoic acid
     (C) o-hydroxybenzoic acid                                         (D) p-toluic acid
5.   Ans.(C)

                                                  1

            O                          O          O
                                              C       H
            C–OH                                             Intramolecular
               OH                                     O
                         ˆˆ†                                 H-bonding
                         ‡ˆˆ

     Most acidic                  Conjugate base


     Due to ortho effect and chelation in its conjugate base, o-hydroxy benzoic acid is most acidic.


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6.   The major product of the following reaction is
                O
             C        (i) KOH
               NH
             C       (ii) Br        CH2Cl
                O
                       O                                                                   O
                   C                                                                   C
     (A)               N–CH2               Br                          (B)                 N             CH2Cl
                   C                                                                   C
                        O                                                                   O

                       O                                                                    O
                   C                                                                   C



6.
     (C)


     Ans.(A)  EN
               O
               C
                       N
                       O–CH2–              Br

                                                          O
                                                          C
                                                                       (D)                 N
                                                                                           O              CH2Cl




                                                                   –    +
                   N–H + KOH                                    N K
            LL
               C                                          C
               O                                          O
                                                                            Br                 CH2 – Cl


                                                                       O
           A

                                                                       C
                                                                             N – CH2                    Br
                                                                       C
                                                                       O

7.   Dissolving 120 g of urea (mol. wt. 60) in 1000 g of water gave a solution of density 1.15 g/mL. The
     molarity of the solution is
     (A) 1.78 M                    (B) 2.00 M                          (C) 2.05 M                       (D) 2.22 M
7.   Ans.(C)
     Mass of solute (urea) = 120 g
     Mass of solvent (water) = 1000 g
     Total mass of solution = (mass of solute + mass of solvent)
                            = 120 + 1000
                            = 1120 g
               2
     M=               ´ 1000 = 2.05 M
           1120 /1.15

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                                        SECTION - II (TOTAL MARKS : 16)
                                  (Multiple Correct Answers Type)
     This section contains 4 multiple choice questions. Each questions has four choices (A), (B), (C) and
     (D) out of which ONE or MORE may be correct.
8.   Extraction of metal from the ore cassiterite involves
     (A) carbon reduction of an oxide ore                              (B) self-reduction of a sulphide ore
     (C) removal of copper impurity                                    (D) removal of iron impurity
8.   Ans.(A,C,D)
     Extraction of tin from cassiterite (SnO2) involves reduction of ore (oxide) by carbon. When the
     concentrated tin stone ore SnO2 (ore of Sn) is heated strongly in a free supply of air (roasting) the
     impurities of CuS and FeS present in the ore are converted into CuSO4 and FeSO4 respectively. The
     CuSO4 and FeSO4 are water soluble which are leached out by hot water.




9.            END
                  ®                  D
     CuS + 2O2 ¾¾ CuSO4 ; FeS + 2O2 ¾¾ FeSO4
                                       ®


     Amongst the given option, the compound(s) in which all the atoms are in one plane in all the possible
     conformations (if any), is (are) -

               H           H                                                                         H
                   C–C
            LL
     (A)                                                               (B) H – C º C – C
           H2C             CH2                                                                       CH2
     (C) H2C = C = O                                                   (D) H2C = C = CH2
9.   Ans.(B,C)
           A

                   H
           H             CH2

     (a)                                 Different conformations don’t have all atoms in the same plane.
                   CH2


                                 CH2
     (b)       H–CºC–C                       Molecular plane
                                 H


               H
     (c)           C=C=O                Molecular plane
           H

           H                   H
     (d)       C=C=C                    All atoms are not present in the same plane.
           H                   H
     Ans. is (B) and (C)
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10.   The correct statement(s) pertaining to the adsorption of a gas on a solid surface is (are) -
      (A) Adsorption is always exothermic
      (B) Physisorption may transform into chemisorption at high temperature
      (C) Physisorption increases with increasing temperature but chemisorption decreases with increasing
          temperature
      (D) Chemisorption is more exothermic than physisorption, however it is very slow due to higher energy
          of activation
10.   Ans.(A,B,D)
      Note : Option (A) is correct as per NCERT book but few examples are known where DHadsorption
      is even endothermic.
11.   According to kinetic theory of gases




11.
             EN
      (A) collisions are always elastic
      (B) heavier molecules transfer more momentum to the wall of the container
      (C) only a small number of molecules have very high velocity
      (D) between collisions, the molecules move in straight lines with constant velocities.
      Ans.(A,B,C,D)
           LL
                                         SECTION - III (TOTAL MARKS : 15)
                                             (Paragraph Type)
      This section contains 2 paragraphs. Based upon one of the paragraph, 3 multiple choice questions
      and based on the other paragraph 2 multiple choice questions have to be answered. Each of these
      questions has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct.
          A

                                 Paragraph for Questions Nos. 12 to 14
      When a metal rod M is dipped into an aqueous colourless concentrated solution of compound N, the
      solution turns light blue. Addition of aqueous NaCl to the blue solution gives a white precipitate O.
      Addition of aqueous NH3 dissolves O and gives in intense blue solution.
12.   The metal rod M is -
      (A) Fe                          (B) Cu                            (C) Ni                             (D) Co
13.   The compound N is -
      (A) AgNO3                       (B) Zn(NO3)2                      (C) Al(NO3)3                       (D) Pb(NO3)2
14.   The final solution contains -
      (A) [Pb(NH3)4]2+ and [CoCl4]2–                                    (B) [Al(NH3)4]3+ and [Cu(NH3)4]2+
      (C) [Ag(NH3)2]+ and [Cu(NH3)4]2+                                  (D) [Ag(NH3)2]+ and [Ni(NH3)6]2+




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      Sol. of 12 to 14
      M - Cu , N - AgNO3 , O - AgCl
      Cu + 2AgNO3 ¾® Cu(NO3)2 + 2Ag + AgNO3 (unreacted)
              excess    light Blue
      AgNO3 + NaCl ¾® AgCl¯ + NaNO3
      (unreacted)     white ppt
      AgCl + 2NH3 ¾® [Ag(NH3)2]+ + Cl–
      Cu(NO3)2 + 4 NH3 ¾® [Cu(NH3)4]2+ + NO3
                                           –


                                   deep Blue
12.   Ans.(B)
13.   Ans.(A)
14.   Ans.(C)



             EN                Paragraph for Questions Nos. 15 to 16
      An acyclic hydrocarbon P, having molecular formula C6H10, gave acetone as the only organic product
      through the following sequence of reactions, in the which Q is an intermediate organic compound.


                                (i) dil H2SO4 /HgSO4
                                                                   (i) conc.H2SO4
                                                                       (Catalytic amount)
                                                                            (–H2O)
                                                                                                               O
                       P                                      Q                                     2          C
           LL
                     (C6H10) (ii) NaBH4/ethanol                    (ii) O3                              H3C        CH3
                                (iii) dil.acid                     (iii) Zn/H2O

15.   The structure of compound P is -
      (A) CH3CH2CH2CH2–CºC–H                                      (B) H3CH2CºC–CH2CH3
          A

         H3C                                                          H3C
      (C) H–C–CºC–CH3                                             (D) H3C–C–CºC–H
         H3C                                                          H3C
16.   The structure of the compound Q is -

         H3C OH                                                        H3C OH
      (A) H–C–C–CH2CH3                                            (B) H3C–C–C–CH3
         H3C H                                                        H3C H

         H3C     OH                                                                      OH
      (C) H–C–CH2CHCH3                                            (D) CH3CH2CH2CHCH2CH3
         H3C




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      Sol. of 15 to 16
15.   Ans.(D)
16.   Ans.(B)
                                                                    HgSO4/dil.H 2SO 4
                                P Þ Me3C – C º CH                                                 Me3C – C – CH3
                                                                                                              O
                                                                                                               NaBH4 / Ethanol
                                                                                                                 dil.acid
                                                         H3 C    +                                    H3C
           H3 C    +                       1,2 shift                                      H2SO4
                                                                                                              C – CH – CH3
                   C – CH – CH 3               –         H3C C – CH – CH 3
           H3 C                             of CH3       H3C                                          H3C
                          CH3                                                                                 CH3 OH
                      +
                –H
                                                          O
      H3 C      CH3               O3/Zn
           C= C                               2           C
      H3 C      CH3


             EN                                   H3 C        CH3

                                        SECTION - IV (TOTAL MARKS : 28)
                                          (Integer Answer Type)
      This section contains 7 questions. The answer to each of the questions is a single-digit integer, ranging
      from 0 to 9. The bubble corresponding to the correct answer is to be darkened in the ORS.
           LL
17.   Reaction of Br2 with Na2CO3 in aqueous solution gives sodium bromide and sodium bromate with
      evolution of CO2 gas. The number of sodium bromide molecules involved in the balanced chemical
      equation is.
17.   Ans.(5)
                                 OH           –
                      disproportionation reaction ®
      3Br2 + 3Na2CO3 ¾¾¾¾¾¾¾¾¾ 5NaBr + NaBrO3 + 3CO2
          A


18.   The difference in the oxidation numbers of the two types of sulphur atoms in Na2S4O6 is.
18.   Ans.(5)

                  O                   O
      Na O — S — S — S — S — O Na
                  O (+5) (0)          O

      Hence, the difference in oxidation states of ‘S’ atom = 5 – 0 = 5
19.   The maximum number of electrons that can have principal quantum number, n=3, and spin quantum
      number, ms = – 1/2, is
19.   Ans.(9)
                                  No. of e– having ms=–1/2
      n=3,          l=0                      1
                    l=1                      3
                    l=2                      5
      Total No. of e– = 1        +3+5 =9

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20.   A decapeptide (Mol. Wt. 796) on complete hydrolysis gives glycine (Mol. Wt. 75), alanine and
      phenylalanine. Glycine contributes 47.0% to the total weight of the hydrolysed products. The number
      of glycine units present in the decapeptide is
20.   Ans.(6)
      No. of peptide linkage = No. of water molecules added for complete hydrolysis.
      =n–1
      So, number of molecules of H2O added = 9
      So total wt. of the product = Mol. wt. of polypeptide + total wt. of H2O added.
      = 796 + (9 × 16)
      = 796 + 162
      = 958
                                         47
\     wt. of glycine obtained = 958 ×        ; 450
                                        100




21.
             EN
      No. of units of glycine =
                                450
                                 75
                                    = 6 units


      To an evacuated vessel with movable piston under external pressure of 1 atm., 0.1 mol of He and 1.0
      mol of an unknown compound (vapour pressure 0.68 atm. at 0°C) are introduced. Considering the
      ideal gas behaviour, the total volume (in litre) of the gases at 0°C is close to
           LL
                                                                                          P =1atm
21.   Ans.(7)
      Since external pressure is 1 atm
      PV + PHe = 1
      Þ PHe = 1 – 0.68 = 0.32 atm
      Now from idal gas equation Þ PV = nRT                                      0.1 mol He
          A

                                                              PV=0.68 atm
      Þ 0.32 × V = 0.1 × (R × 273)                                               1 mol X(s)
      Þ V = 7 litre.


22.   The total number of alkenes possible by dehydrobromination of 3-bromo-3-cyclopentylhexane using
      alcoholic KOH is
22.   Ans.(5)

          Br

  C–C–C–C–C–C                       C–C–C–C–C–C                          C–C=C–C–C–C                          C–C–C=C–C–C
                       alcoholic
                                                                    +                                    +
                        KOH
                                            1 product                        2 products                           2 products
                                                                         (Geometrical isomers)                (Geometrical isomers)




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23.   The work function (f) of some metals is listed below. The number of metals which will show photoelectric
      effect when light of 300 nm wavelength falls on the metal is : :-

          Metal Li Na K Mg Cu Ag Fe Pt           W
          f(eV) 2.4 2.3 2.2 3.7 4.8 4.3 4.7 6.3 4.75

23.   Ans.(4)
      l = 300 nm

           1240
      E = l in nm eV

          1240
      =        eV
           300
      = 4.13 eV


               EN
      To show photoelectric effect E ³ f.
      Total no. of metals that show photoelectric effect will be 4.
             LL
            A




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                                                      PART - II (PHYSICS)
                                           SECTION–I : (Total Marks : 21)
                                             (Single Correct Answer Type)
      This section contains 7 multiple choice questions. Each question has four choices (A), (B), (C) and
      (D), out of which ONLY ONE is correct.

24.   5.6 liter of helium gas at STP is adiabatically compressed to 0.7 liter. Taking the initial temperature to be
      T1, the work done in the process is
            9                              3                             15                                    9
      (A)     RT                 (B)         RT                    (C)      RT                        (D)        RT
            8 1                            2 1                            8 1                                  2 1
Ans. (A)
                            5.6 1
      No. of moles = n =        =
                            22.4 4
               EN
      TVg–1= constant ÞT1 (5.6)2/3 = T2 (0.7)2/3 Þ T1(8)2/3 = T2 Þ 4T1 = T2

      W=
            - nRDT
              g -1
                   =-
                      1R ( 3T1 ) ´ 3
                         4´2
                                        9
                                     = - RT1 . Therefore W
                                        8                  external
                                                                     9
                                                                    = RT1
                                                                     8
                                                                                                                                             \\\\\\\\\\\\\\
25.  A ball of mass (m) 0.5 kg is attached to the end of a string having length (L) 0.5 m.
      The ball is rotated on a horizontal circular path about vertical axis. The maximum                                                 L
             LL
     tension that the string can bear is 324 N. The maximum possible value of angular
     velocity of ball (in radian/s) is
                                                                                                                            m
     (A) 9                      (B) 18                   (C) 27                   (D) 36
Ans. (D)
     Tsinq=mw2r                        \\\\\\\\\\\\\\

     Tsinq=mw Lsinq
                 2
            A


                                           q
     T = mw2L                       q

              1 ( 2)1
      324 =      w                                q
              2     2
      Therefore w = 36
                                   r                                                                                    z
26.   Consider an electric field       =     , where E0 is a constant. The flux through
      the shaded area (as shown in the figure) due to this field is                                        (a,0,a))         (a,a,a)

      (A) 2E0a2                                                    (B)     2E0 a 2

                                                                         E 0 a2                                         (0,0,0)       (0,a,0)
                                                                                                                                                         y
      (C) E0a 2
                                                                   (D)
                                                                             2                             x

Ans. (C)
               uur
      f = ò E .dS = Ex projected area perpendiuclar to E (x-axis) = E × a2


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27.  A police car with a siren of frequency 8 kHz is moving with uniform velocity 36 km/hr towards a tall
     building which reflects the sound waves. The speed of sound in air is 320 m/s. The frequency of the
     siren heard by the car driver is
     (A) 8.50 kHz             (B) 8.25 kHz           (C) 7.75 kHz           (D) 7.50 kHz
Ans. (A)
         æ v ö æ v + v0 ö      æ 320 ö æ 320 + 10 ö
      f¢=ç        ÷ç    ÷ f f¢=ç                    ´ 8 Þ f ¢ » 8.50 kHz
         è v - vs ø è v ø Þ    è 320 - 10 ÷ ç 320 ÷
                                          øè      ø
28.   A meter bridge is set-up as shown, to determine an unknown resistance ‘X’ using a standard 10 ohm
      resistor. The galvanometer shows null point when tapping-key is at 52 cm mark. The end-corrections
      are 1 cm and 2 cm respectively for the ends A and B. The determined value of ‘X’ is

                                                             X                   W




                                                     A                                      B




Ans. (B)
             EN
     (A) 10.2 ohm                     (B) 10.6 ohm

                                                x
                                                    =
                                                        10          10
                                                                       (C) 10.8 ohm                        (D) 11.1 ohm


      Apply condition of wheatstone bridge,                  Þ x=        ´ 53 Þx = 10.6 W
                                             52 + 1 48 + 2           50
29.   A 2 mF capacitor is charged as shown in figure. The percentage of its                                                      1    2
           LL
                                                                                                                                      s
      stored energy dissipated after the switch S is turned to position 2 is
                                                                                                                       V
                                                                                                                               2m F         8m F
      (A) 0%                                                           (B) 20%

     (C) 75%                                                           (D) 80%
Ans. (D)
          A


                                                                                 -
                  - -- - -                                              -----           -----
                2m                   m                                 2m               m




                                                                                                                                2
                                                                            æ 2V ö
                         2V - x x     8V       1                         2  ç ÷
                                         Þ Vi = ´ ( 2) V = V ; U = æ 8V ö + è 5 ø = 4V
                                                                                       2
      Q1 = CV , Q1 = 2V,       = , x=                   2   2

                           2    8      5       2                f  ç ÷
                                                                   è 5ø      2´2     5

                                         4V 2
             4V 2                             ´ 100
      Loss =      Þ % loss =              5         = 80%
              5                              V2



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30.  The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 Å. The wavelength
     of the second spectral line in the Balmer series of singly-ionized helium atom is
     (A) 1215 Å               (B) 1640Å                 (C) 2430Å                 (D) 4687Å
Ans. (A)

      1        æ 1   1 ö
        = Rz 2 ç 2 - 2 ÷
      l        èn m ø

                                                       1          2 æ 1  1ö
      First line of Balmer of Hydrogen :                   = R (1) ç 2 - 2 ÷
                                                      6561          è2 3 ø

                                                                      1           æ 1 1ö
      Second line of Balmer of single ionized He :                      = R (z2 ) ç 2 - 2 ÷
                                                                      l           è2   4 ø

                                     5
      Dividing : l = 6561 ´             = 1215 Å
                                    9´3
             EN                               SECTION–II : (Total Marks : 16)
                                               (Multiple Correct Answer Type)
      This section contains 4 multiple choice questions. Each question has four choices (A), (B), (C) and
      (D), out of which ONE or MORE may be correct.
31.   A spherical metal shell A of radius RA and a solid metal sphere B of radius RB (<RA) are kept far apart
      and each is given charge ‘+Q’. Now they are connected by a thin metal wire. Then
           LL
                                                                              s
      (A) E A = 0
            inside
                                      (B) QA > QB                       (C) s =                            (D) E A surface < EB surface
                                                                                                                 on           on



Ans. (ABCD)
                                                                  Q                                            QB
                                     Q                                            QA
          A


                                     RB                            RA             RB                           RA


             =    (because of electrostatic condition) So, A option is true.

                  kQA kQB Q A RA
      Þ vA = vB Þ R = R Þ Q = R Þ RB < RA So, QB < QA, so B is true
                    A   B   B  B



        s A 4 pRA R A
                2
                        s A RB
      Þ           =   Þ s =
        s B 4pRB RB
                2
                          B RA , So C is true

                            1
      Enear surface = s ×     . So, D is also true
                            R




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32.  A metal rod of length ‘L’ and mass ‘m’ is pivoted at one end. A thin disk of mass ‘M’ and radius ‘R’
     (<L) is attached at its center to the free end of the rod. Consider two ways the disc is attached: (case A).
     The disc is not free to rotate about its center and (case B) the disc is free to rotated about its center. The
     rod-disc system performs SHM in vertical plane after being released from the same displaced position.
     Which of the following statement(s) is(are) true?
     (A) Restoring torque in case A = Restoring torque in case B
     (B) Restoring torque in case A < Restoring torque in case B
     (C) Angular frequency for case A> Angular frequency for case B
     (D) Angular frequency for case A< Angular frequency for case B
Ans. (AD)
     Torque for both the arrangement is same.
     Since in case B disc is not rotating, there is no speed of the pendulum at equilibrium in case (B).
33. An electron and a proton are moving on straight parallel paths with same velocity. They enter a semi-
     infinite region of uniform magnetic field perpendicular to the velocity. Which of the following statement(s)
     is/are true?
     (A) they will never come out of the magnetic field region
     (B) they will come out travelling along parallel paths
                EN
     (C) they will come out of the same time
     (D) they will come out at different times
Ans. (BD)
                               By diagram B is true.
                                            2 pm
                                     T=
         p                                   qB
              LL
         e
                                Tµm
                                mP > m e
                                Tp > Te
                                So, D is also true.
34. A composite block is made of slabs A, B, C, D and E of different thermal conductivities (given in terms
             A


     of a constant K) and sizes (given in terms of length, L) as shown in the figure. All slabs are of same
     width. Heat ‘Q’ flows only from left to right through the blocks. Then in steady state
     (A) heat flow through A and E slabs are same                           heat     0 1L                5L 6L
     (B) heat flow through slab E is maximum                                      1L
                                                                                        A        B 3K        E
     (C) temperature difference across slab E is smallest                              2K        C 4K       6K
     (D) heat flow through C= heat flow through B + Heat flow through D           3L
                                                                                                 D 5K
Ans. (ABCD)                                                                       4L
     • In steady state : heat in = heat out. So, A is true
     • Option B is also true because total heat is flowing through E.                     B 4R/3

             DT                                                                    A      C R/2        R/24
     • Q=
              R                                                                   R/8                    E
     Q = same                                                                             D 4R/5
     RE is minimum. So, DT is minimum
     So option C is true
                                                                                    R/8    R/4    R/24
               DT             DT             DT
     • QB =         ,Q =            , QD =         , So, QB + QD= QC.
              4R / 3 C 4 R / 2              4R / 5
     Hence D is also true.
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                                 SECTION–III : (Total Marks : 15)
                                           (Paragraph Type)
      This section contains 2 paragraphs. Based upon one of the paragraph, 3 multiple choice questions
      and based on the other paragraph 2 multiple choice questions have to be answered. Each of these
      questions has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct.
                                   Paragraph for Questions Nos. 35 to 37
      Phase space diagrams are useful tools in analyzing all kinds of dynamical problems. They are especially
      useful in studying the changes in motion as initial position and momentum are changed. Here we consider
      some simple dynamical systems in one-dimension. For such systems, phase space is a plane in which
      position is plotted along horizontal axis and momentum is plotted along vertical axis. The phase space
      diagram is x(t) vs. p(t) curve in this plane. The arrow on the curve indicates the time flow. For example,
      the phase space diagram for a particle moving with constant velocity is a straight line as shown in the
      figure. We use the sign convention in which position or momentum upwards (or to right) is positive and
      downwards (or to left) is negative.
                                              Momentum




               EN
                                                                                      Position

35.   The phase space diagram for a ball thrown vertically up from ground is
             LL
                 Momentum                                Momentum                         Momentum                        Momentum




      (A)                   position   (B)                          position   (C)                    position   (D)                 position
            A


Ans. (D)
     Initial momentum was positive and final momentum negative. So option (D) is correct.
36. The phase space diagram for simple harmonic motion is a circle centered at the origin. In the figure, the
     two circles represent the same oscillator but for different initial conditions, and E1 and E2 are the total
     mechanical energies respectively. Then
                                                                     momentum



                                                                                     E1
                                                                                E2

                                                                                     2a

                                                                                a          position




      (A)    =                         (B) E1=2E2                          (C) E1=4E2                            (D) E1=16E2
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Ans. (C)
                                                    2
                           E2 æ a ö
      E µ (amplitude) Þ so2  = ç ÷ Þ E1 = 4 E2
                           E1 è 2 a ø
37.   Consider the spring-mass system, with the mass submerged in water, as shown in the figure. The phase
      space diagram for one cycle of this system is
                                                                    \\\\\\\\\\\




                   momentum                                                            momentum




      (A)      EN  momentum
                                position
                                                                        (B)

                                                                                       momentum
                                                                                                      position




      (C)                                                               (D)
             LL
                                position                                                              position




Ans. (B)
     Since at start time position was positive
                                    Paragraph for Question Nos. 38 and 39
            A


      A dense collection of equal number of electrons and positive ions is called neutral plasma. Certain solids
      containing fixed positive ions surrounded by free electrons can be treated as neutral plasma. Let ‘N’ be
      the number density of free electrons, each of mass ‘m’. When the electrons are subjected to an electric
      field, they are displaced relatively away from the heavy positive ions. If the electric field becomes zero,
      the electrons begins to oscillate about the positive ions with a natural angular frequency ‘wr’, which is
      called the plasma frequency. To sustain the oscillations, a time varying electric field needs to be applied
      that has an angular frequency w, where a part of the energy is absorbed and a part of it is reflected. As w
      approaches wr, all the free electrons are set to resonance together and all the energy is reflected. This is
      the explanation of high reflectivity of metals.




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38.   Taking the electronic charge as ‘e’ and the permittivity as ‘e0’, use dimensional analysis to determine the
      correct expression for wr.


             Ne                            me 0                              Ne 2                               me 0
      (A)                         (B)                               (C)                                (D)
             me 0                          Ne                                me 0                               Ne 2

Ans. (C)


                      æ 1 ö ( 2)
      é Ne 2     ù    ç 3÷ C
                      èL ø          1
      ê          ú=                = = [w]
      ê m Î0              æ      ö T
                             2 2
      ë          ú
                 û  ( M ) ç C3 T ÷
                          è LMø


39.            EN
      Estimate the wavelength at which plasma reflection will occur for a metal having the density of electrons
                                        -11
      N » 4 ´ 10 27 m -3 . Take e 0 » 10 and m » 10 -30 , where these quantities are in proper SI units

      (A) 800 nm                  (B) 600 nm                        (C) 300 nm                         (D) 200 nm
Ans. (B)
             LL
                     2 pc   Ne 2            m Î0
      w = 2 pf =          =      Þ l = 2 pc
                      l     m Î0            Ne2


            2 ´ 3.14 ´ 3 ´ 108    (10 -30 )(10 -11 ) 9.42
      l=                                            =     ´ 10 27 ´ 10 -34 = 6 ´ 10 -7 m = 600 nm
               1.6 ´ 10 -19         ( 4 ´ 10 27 )     1.6
            A


                                         SECTION–IV : (Total Marks : 28)
                                                   (Integer Answer Type)
      This Section contains 7 questions. The answer to each of the question is a single digit integer, ranging
      from 0 to 9. The bubble corresponding to the correct answer is to be darkened in the ORS.
40.  A block is moving on an inclined plane making an angle 45° with the horizontal and the coefficient of
     friction is m. The force required to just push it up the inclined plane is 3 times the force required to just
     prevent it from sliding down. If we define N =10m, then N is
Ans. (5)
     Force to just prevent it from sliding = mgsinq – mmgcosq
     Force to just push up the plane = mgsinq + mmgcosq
     mgsinq + mmgcosq = 3 (mgsinq – mmgcosq)

       1         1    æ 1   m ö       1
            +m     = 3ç   -    ÷ Þ m = ÞN = 10 m =5
        2        2    è 2    2 ø      2

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                                                                                                                                       stick
41.  A boy is pushing a ring of mass 2 kg and radius 0.5 m with a stick as
     shown in the figure. The stick applies a force of 2N on the ring and rolls
     it without slipping with an acceleration of 0.3 m/s2. The coefficient of
     friction between the ground and the ring is large enough that rolling always
     occurs and the coefficient of friction between the stick and the ring is (P/
     10). The value of P is                                                                                                   Ground
Ans. (4)
     N1 = 2 N
                                                                                  mN1
     N1 – f = ma ...(i)
     (f–mN1)R = mR2a = ma ...(ii)
                                                           a
     From equation (i) and (ii) we get                                    W           N1
     N1 ( 1–m) = 2ma
     2 (1–m) = 2 × 2 × 0.3                                                      f
     1 – m = 0.6 Þ m = 0.4

42.
                  EN
      Four point charges, each of +q, are rigidly fixed at the four corners of a square planar soap film of side
      ‘a’. The surface tension of the soap film is g. The system of charges and planar film are in equilibrium,
                            1/ N
             é q2 ù
      and a=kê ú                    , where ‘k’ is a constant. Then N is
             ëg û
Ans. (3)
                LL

                                         b
           q                        q


                                             Line ab divides the soap film into two equal parts.
               A


            q                 q
      a

                 FBD of half part


          gÖ2a
                                        Kq 2 é    1ù            1
                                          2 ê
                                               2 + ú where K=
                                         a ë      2û          4p Î0


                                                                                     1/ 3                     1/ 3
            Kq 2 æ  1ö       Kq 2 æ   1ö    é q2 ù                                          é æ      1ö ù
      g 2a = 2 ç 2 + ÷ ; a =      ç 2+ ÷; a=ê ú
                          3

                              g è     2ø                                                    êK ç 2 + 2 ÷ ú           ÞN=3
             a è    2ø                      ëg û                                            ë è        øû

                                         1/ 3
            é æ      1ö ù
      where ê K ç 2 + ÷ ú
                è
                                                =k
            ë        2ø û


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43.   Four solid spheres each of diameter    5 cm and mass 0.5 kg are placed with their centers at the
     corners of a square of side 4 cm. The moment of inertia of the system about the diagonal of the square is
     N × 10–4 kg-m2, then N is
Ans. (9)
                                                              2                              2
         2     2     2        æ a ö   2        æ a ö
                                  ÷ + 5 mR + m ç
                                          2
      I = mR2 + mR2 + mR2 + m ç
         5     5     5        è 2ø             è 2÷ø

           8               é8       5            ù
      I=     mR 2 + ma 2 = ê ´ 0.5 ´ + 0.5 ´ 4 2 ú ´ 10 -4 = (1+8) × 10–4 = N × 10–4 ÞN =9
           5               ë5       4            û
44.  The activity of a freshly prepared radioactive sample is 1010 disintegrations per second, whose mean life is
     109 s. The mass of an atom of this radioisotope is 10–25 kg. The mass (in mg) of the radioactive sample is
Ans. (1)

                             1010
      A = lN Þ 10 = lN Þ N =
                       10         = (1010 ) t = 1010 ´ 10 9 = 1019
              EN              l
      M = Nm = (1019) (10–25) = 10–6 kg = 1 mg
45.   A long circular tube of length 10 m and radius 0.3 m carries a current I along its
      curved surface as shown. A wire-loop of resistance 0.005 ohm and of radius
      0.1 m is placed inside the tube with its axis coinciding with the axis of the tube.
      The current varies as I=I0cos (300t) where I0 is constant. If the magnetic moment
            LL
      of the loop is N m 0 I 0 sin ( 300t ) , then ‘N’ is
Ans. (6)

                æ m0 I ö 2            pr 2                d f æ m 0 I 0 pr 2 ö
      f = Bpr = ç        pr = m 0 I 0      cos300t Þ e1 =    =ç              ÷ 300 sin 300t
                è L ÷
               2
                       ø               L                  dt è L ø
           A


         e                        é pr 2 (300 ) ù                é p 2r 4 ( 300 ) ù
      i=   = ( m 0 I 0 sin 300t ) ê               Þ M = i pr 2 = ê                ú m 0 I 0 sin 300t
         R                        ë LR ú        û                ë      LR        û
46.  Steel wire of length ‘L’ at 40°C is suspended from the ceiling and then a mass ‘m’ is hung from its free
     end. The wire is cooled down from 40° to 30° C to regain its original length ‘L’. The coefficient of linear
     thermal expansion of the steel is 10–5/°C, Young’s modulus of steel is 1011 N/m2 and radius of the wire
     is 1 mm. Assume that L>> diameter of the wire. Then the value of ‘m’ in kg is nearly
Ans. (3)
      Dx YA              YA
        =   = aDq Þ m =
      L mg              gaDq

            (1011 ) ( 3.14) (10 -6 )
      m=                             Þ m = 3.14kg Þ m = 3
              (10 ) (10 -5 ) (10 )




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                                                  PART - III (MATHEMATICS)
                                               SECTION–I : (Total Marks : 21)
                                                      (Single Correct Answer Type)
      This section contains 7 multiple choice questions. Each question has four choices (A), (B), (C)
      and (D), out of which ONLY ONE is correct.

                          ln 3
                                    x sin x 2
47.   The value of        ò sin x 2 + sin(ln 6 - x 2 ) dx is
                          ln 2


          1 3                                 1 3                                        3                                1 3
      (A)   ln                         (B)     ln                           (C) ln                               (D)       ln
          4 2                                 2 2                                        2                                6 2
47.   Ans.(A)

      I= ò



      Þ
         ln2
            ln3




              1
               EN
             sin x 2 + sin(ln6 - x 2 ) ; put x = t Þ 2xdx = dt


            I= ò
                  ln3
                     x sin x 2


                             sin t
                                      dx



              2 ln 2 sin t + sin(ln6 - t)
                  ln3
                                          dt
                                              2




                                                             ....(i)

             1         sin(ln6 - t)
      Þ I= ò
             LL
                                         dt                  ....(ii)
             2 ln 2 sin(ln6 - t) + sin t
      Adding equation (i) & (ii)
                    ln3
                1                 1 æ3ö
      Þ 2I = ò dt Þ I = ln ç ÷
                2 ln2             4 è2ø
      Let the straight line x = b divide the area enclosed by y = (1 – x)2, y = 0 and x = 0 into two parts
            A

48.
                                                                                         1
      R1 (0 £ x £ b) d and R 2 (b £ x £ 1) such that R1 - R 2 =                            . Then b equals
                                                                                         4
          3                                   1                                      1                                    1
      (A)                              (B)                                  (C)                                  (D)
          4                                   2                                      3                                    4
                                                                    y
48.   Ans.(B)
                          1
      Q     R1 –R2 =
                          4
            b                     1                                R1 R
                                                                      R2
                                                       1
            ò (1 - x) dx - ò (1 - x) dx =                                                         x
                                                                            2
                     2              2
      Þ                                                           0 x=b          (1,0)
            0                     b
                                                       4
                              b                   1
             æ (1 - x)3 ö æ (1 - x) 3 ö   1                             ì (1 - b)3 1 ü (1 - b)3 1
      Þ     -ç          ÷ +ç          ÷ =                    Þ         -í         - ý-         =
             è 3 ø0 è 3 ø b 4                                           î 3        3þ      3     4
            1 2           1                    2             1
      Þ      - (1 - b)3 =   Þ                    (1 - b)3 =
            3 3           4                    3            12
                              1                            1                         1
      Þ (1 – b)3 =                     Þ 1–b=                       Þ           b=
                              8                            2                         2

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                         r                                                          r                 r     r
      Let a = ˆ + ˆ + k, b = ˆ - ˆ + k and r = ˆ - ˆ - k be three vectors. A vector v in the plane of a and b ,
          r
49.           i j ˆ          i j ˆ         c i j ˆ

                          r    1
      whose projection on c is    , is given by
                                3
      (A) ˆ - 3ˆ + 3k
            i  j     ˆ            ˆ
                            (B) -3i - 3ˆ - k
                                       j ˆ                                     (C) 3i - ˆ + 3k
                                                                                    ˆ j      ˆ                      (D) ˆ + 3ˆ - 3k
                                                                                                                        i    j ˆ
49.   Ans.(C)
       r     r r
      v = xa + yb
      = ˆ + y) + ˆ - y) + k(x + y) ....(i)
         i(x       j(x    ˆ

              r     1
      Given , v.c =
                ˆ
                     3
           x+y-x+y-x-y               1
      Þ                         =




50.
          y–x=1
               EN
      Þ x – y = –1
                      3

                              ....(ii)
                                r
                                      3


                                          ˆ j
      using (ii) in (i) we get v = (x + y)i - ˆ + (x + y)k
                                                         ˆ
      Let (x0, y0) be the solution of the following equations
                                          ( 2x )
                                                   ln 2
                                                          = (3y)ln 3
             LL
                                          3lnx = 2lny
      Then x0 is
            1                                    1                                     1
      (A)                                 (B)                                  (C)                                  (D) 6
            6                                    3                                     2
50.   Ans.(C)
                     ln2
            ( 2x )
            A

                           = (3y) l n 3
      Þ ln2 (ln2 + lnx) = ln3(ln3 + lny) .....(i)
        3lnx = 2lny
      Þ (lnx) (ln3) = (lny) (ln2)        .....(ii)
        using (ii) in (i)
                          æ       (lnx)(ln3) ö
      Þ ln2(ln2+lnx)= ln3 ç ln3 +            ÷
                          è          ln2     ø
                              ì ln 2 3       ü
      Þ ln 2 2 - ln 2 3 = lnx í        - ln2 ý
                              î ln2          þ
      Þ lnx = – ln2
                 1
      Þ     x=
                 2

51.   Let a and b be the roots of x2 – 6x – 2 = 0, with a > b. If an = an – bn for n ³ 1, then the value
         a10 - 2a 8
      of    2a 9
                    is
      (A) 1                               (B) 2                                (C) 3                                (D) 4

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51.   Ans. (C)
      a,b are roots of x2 – 6x – 2 = 0
      Þ a2 – 6a –2 = 0 & b2 – 6b – 2 = 0

            a 0 - 2a 8 a10 - b10 - 2(a8 - b8 )
                      =
               2a 9         2(a 9 - b9 )

                a8 (a2 - 2) - b8 (b2 - 2)
            =
                      2(a9 - b9 )

                a8 .6a - b8 .6b
            =                   =3
                 2(a9 - b9 )

52.   A straight line L through the point (3, –2) is inclined at an angle 60° to the line                                              3x + y = 1 . If L




52.
      also intersect the x-axis, then the equation of L is
      (A) y + 3x + 2 - 3 3 = 0

      (C)
      Ans. (B)
               EN
             3y - x + 3 + 2 3 = 0
                                                                               (B) y - 3x + 2 + 3 3 = 0

                                                                               (D)       3y + x - 3 + 2 3 = 0

                                                                                                                         (0,1)
                                                          p
             LL
      Line L has two possible slopes with inclination; q = , q = 0
                                                          3

                                                             p                                                                                  2p/3
      \     equation of line L when q =                        ,        y+2=          3(x - 3)
                                                             3
                                                                                                                                  GH   1
                                                                                                                                        3
                                                                                                                                          ,0   JK      3x+y=1
                                                                   Þ    y - 3x + 2 + 3 3 = 0
            A

            equation of line L when q = 0,                             y = –2 (rejected)

      \     required line L is y - 3x + 2 + 3 3 = 0

53.              {                                       }                 {
      Let P = q : sin q - cos q = 2 cos q and Q = q : sin q + cos q = 2 sin q be two sets. Then                   }
      (A) P Ì Q and Q - P ¹ Æ                                                  (B) Q Ì P
                                                                                     /
      (C) P Ì Q
            /                                                                  (D) P = Q
53.   Ans.(D)
      P = {q : sinq – cosq =                   2 cosq}
      Þ tanq =             2 +1                          ...(i)

      Q = {q : sinq + cos q =                     2 sinq}
                            1
      Þ     tan q =             = 2 +1                   ...(ii)
                           2 -1
      from (i) & (ii)
      Þ     P=Q
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                                SECTION–II : (Total Marks : 16)
                                  (Multiple Correct Answer Type)
      This section contains 4 multiple choice questions. Each question has four choices (A), (B), (C)
      and (D), out of which ONE or MORE may be correct.
54.                                                             ˆ          j ˆ
      The vector(s) which is/are coplanar with vectors ˆ + ˆ + 2k and ˆ + 2ˆ + k , and perpendicular to the
                                                       i j            i
             i j ˆ
      vector ˆ + ˆ + k is/are

          j ˆ
      (A) ˆ - k                    (B) -ˆ + ˆ
                                        i j                             (C) ˆ - ˆ
                                                                            i j                                        ˆ
                                                                                                             (D) -ˆj + k
Sol. Ans. (A,D)
      r                     r                             r
      a = ˆ + ˆ + 2k
          i j      ˆ        b = ˆ + 2ˆ + k
                                i     j ˆ                     i j ˆ
                                                          c = ˆ+ˆ+k
      r       r r r           rr r rr r
      v = l((a ´ b) ´ c) = l((a.c)b - (b.c)a
      r



55.
      r
             EN
      v = l[4(i + 2 ˆ + k) - 4(i + ˆ + 2k)]
                ˆ

      v = 4 l(ˆ - k)
              j ˆ
                     j ˆ       ˆ j      ˆ



     Let ƒ : R ® R be a function such that
                              ƒ(x + y) = ƒ(x) + ƒ(y), " x, y Î R.
     If ƒ(x) is differentiable at x = 0, then
     (A) ƒ(x) is differentiable only in a finite interval containing zero
           LL
     (B) ƒ(x) is continuous " x Î R
     (C) ƒ'(x) is constant " x Î R
     (D) ƒ(x) is differentiable except at finitely many points
Sol. Ans. (B,C)
     ƒ(x + y) = ƒ(x) + ƒ(y)
          A


     ƒ(0) = 0
                     ƒ(x + h) - ƒ(x)        ƒ(x) + ƒ(h) - ƒ(x)
      ƒ'(x) = lim                    = lim
              h ®0          h          h ®0         h
           ƒ(h)        ƒ(0 + h) - ƒ(0)
      = lim     = lim
        h ®0 h    h ®0        h
     ƒ'(x) = ƒ'(0) = k (k is constant)
     Þ ƒ(x) = kx, hence ƒ(x) is continuous and ƒ'(x) is constant " x Î R
56. Let M and N be two 3 × 3 non-singular skew-symmetric matrices such that MN = NM. If PT denotes
     the transpose of P, then M2N2(MTN)–1 (MN–1)T is equal to -
     (A) M2                    (B) –N2           (C) –M2               (D) MN
Sol. Ans. (C)
      (Comment : Although 3 × 3 skew symmetric matrices can never be non-singular. Therefore the
      information given in question is wrong. Now if we consider only non singular skew symmetric
      matrices M & N, then the solution is-)
      Given MT = –M
               NT = –N
               MN = NM
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      according to question M2N2(MTN)–1 (MN–1)T
      = M2N2N–1(MT)–1(N–1)TMT                                                        é MN = NM
          2    2   –1
      = M N N (–M) (N ) (–M)–1     T –1                                              ê       -1         -1
                                                                                     ê (MN) = (NM)
      = –M2 N M–1 N–1 M                                                              ê N -1M -1 = M -1N -1
                                                                                     ë
      = – M2 N N–1 M–1 M = –M2
                                            x 2 y2
57.   Let the eccentricity of the hyperbola 2 - 2 = 1 be reciprocal to that of the ellipse x2 + 4y2 = 4.
                                            a   b
      If the hyperbola passes through a focus of the ellipse, then -
                                                                x2 y 2
      (A) the equation of the hyperbola is                        -    =1
                                                                3 2
      (B) a focus of the hyperbola is (2,0)




             EN
      (C) the eccentricity of the hyperbola is
                                           2
                                              5
                                              3
     (D) the equation of the hyperbola is x – 3y2 = 3
Sol. Ans. (B,D)


                                x 2 y2
      Given        hyperbola is 2 - 2 = 1
           LL
                                a   b
                                  x 2 y2
                   ellipse is        +   =1
                                  22 1
                                                  1         3    3
      eccentricity of ellipse =              1-     =         =
                                                  4         4   2
          A


                                                        b2          4
      eccentricity of hyperbola =                  1+      =
                                                        a2          3

              b2 1
      Þ         =        Þ 3b2 = a2                           ...........(1)
              a2 3
      also hyperbola passes through foci of ellipse ( ± 3, 0)
      3
         =1        Þ a2 = 3                                   ............(2)
      a2
      from (1) & (2)
      b2 = 1
                                             x 2 y2
      equation of hyperbola is                  -   =1                  Þ x2 – 3y2 = 3
                                             3 1
                                                       1        4
      eccentricity of hyperbola = 1 +                    =
                                                       3        3

                           æ     2    ö
      focus of hyperbola = ç ± 3. , 0 ÷ º ( ±2, 0 )
                           è      3 ø
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                                  SECTION–III : (Total Marks : 15)
                                           (Paragraph Type)
       This section contains 2 paragraphs. Based upon one of the paragraph, 3 multiple choice questions
       and based on the other paragraph 2 multiple choice questions have to be answered. Each of these
       questions has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct.
                                  Paragraph for Question 58 and 60
       Let a,b and c be three real numbers satisfying
                  é1 9 7ù
       [ a b c ] ê8 2 7 ú = [ 0 0 0 ]
                  ê          ú                                                    ...(E)
                  ê 7 3 7ú
                  ë          û
58.    If the point P(a,b,c), with reference to (E), lies on the plane 2x + y + z = 1, then the value of 7a+b+c
       is

59.




60.
       (A) 0


       value of
       (A) –2
                EN
       Let w be a solution of x – 1 = 0 with Im(w) > 0. If a = 2 with b and c satisfying (E), then the
                 3
                    + b + c is equal to -
                w w w
                  a
                      1    3
                                (B) 12




                             (B) 2
                                  3




                                                     (C) 3
                                                          (C) 7




                                                                            (D) –3
                                                                                  (D) 6




       Let b = 6, with a and c satisfying (E). If a and b are the roots of the quadratic equation
              LL
                                                     n
                                  æ1 1ö
                                    ¥

       ax + bx + c = 0, then å ç + ÷ is-
         2

                             n= 0 è a bø
                                                         6
       (A) 6                     (B) 7                                (C)                                  (D) ¥
                                                         7
Sol.                              Paragraph for Question 58 to 60
             A

       a + 8b + 7c = 0
       9a + 2b + 3c = 0
       7a + 7b + 7c = 0
       Þ a = K, b = 6K , c = –7K
58.    Ans. (D)
       (K, 6K, –7K)
       2x + y + z = 1
       2K + 6K – 7K = 1       (Q point lies on the plane)
       Þ K=1
       Þ 7a + b + c = 7K + 6K – 7K = 6
59.    Ans. (A)
       x3 – 1 = 0
       Þ x = 1, w, w2
             1 i 3
       w= - +         since Im(w) > 0
             2   2
       If a = 2 = K Þ b = 12 & c = –14

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             3     1   3   3   1    3
      Hence      + b + c = 2 + 12 + -14 = 3w+1+3w2
             w w w w w
               a
                                   w
      = –3 + 1 = –2
60.   Ans. (B)
      Q b = 6 Þ 6K = 6 Þ K = 1
      Þ a = 1,        b = 6 & c = –7
       2
      x + 6x – 7 = 0
      Þ a + b = –6 , ab = –7
                            n                n
              æ a +b ö
               ¥          ¥
                              æ6ö     1
          å ç ab ÷ = å ç 7 ÷ = 6 = 7
     Þ n =0 è            n =0 è ø 1-
                     ø
                                        7
Sol.                            Paragraph for Question 61 and 62




      (A)
          13
               EN
     Let U1 and U2 be two urns such that U1 contains 3 white and 2 red balls, and U2 contains only 1
     white ball. A fair coin is tossed. If head appears then 1 ball is drawn at random from U1 and put
     into U2. However, if tail appears then 2 balls are drawn at random from U1 and put into U2. Now
     1 ball is drawn at random from U2.
61. The probability of the drawn ball from U2 being white is -

                            (B)
                                 23
                                                   (C)
                                                       19
                                                                          (D)
                                                                               11
          30                     30                    30                      30
             LL
62.   Given that the drawn ball from U2 is white, the probability that head appeared on the coin is -
            17                                   11                  15                                                    12
      (A)                               (B)                     (C)                                               (D)
            23                                   23                  23                                                    23
                                                 Paragraph for Question 61 and 62
                          U1                     U2
            A

                          3W
                          2R 1 ball           1W
              1/2 d
                   a
                 He
      Start
                1         U1                     U2
            Ta /2
              il          3W
                                              1W
                          2R 2 balls
61.   Ans. (B)
      Required probability
              2 1 ö 1 æ 3C        C 1 3C C 2 ö
                                2          2
       1æ 3
      = ç .1 + . ÷ + ç 5 2 .1 + 5 2 . + 5 1 1 . ÷
       2è 5   5 2 ø 2 è C2        C2 3    C2 3 ø
        1 æ 4 ö 1 æ 3 1 2 ö 2 11 23
      =   ç ÷+ ç + + ÷ = +        =
        2 è 5 ø 2 è 10 30 5 ø 5 30 30
62.   Ans. (D)
      Required probability
                2/5
      =                  (using Baye's theorem)
          2 / 5 + 11/ 30
          12
      =
          23

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                                               SECTION–IV : (Total Marks : 28)
                                                          (Integer Answer Type)
      This Section contains 7 questions. The answer to each of the question is a single digit integer, ranging
      from 0 to 9. The bubble corresponding to the correct answer is to be darkened in the ORS.
                                                                                                                        p

63.   Let a1,a2,a3,.........,a100 be an arithmetic progression with a1 = 3 and Sp = å a i ,1 £ p £ 100 . For any
                                                                                                                       i =1

                                                Sm
      integer n with 1 < n < 20, let m = 5n. If S does not depend on n, then a2 is
                                                 n

Sol. Ans. 9 or 3
     (Comment : The information about the common difference i.e. zero or non-zero is not given in
     the question. Hence there are two possible answers)




      a1 = 3 ;

      m = 5n
               EN
     Consider d ¹ 0 the solution is
     a1, a2, a3, ............, a100 ® AP
                            p
                    Sp = å a i
                           i =1
                                          1 £ n £ 20



           m
             LL
             [2a1 + (m - 1)d]
      Sm
         = 2
      Sn   n
             [2a1 + (n - 1)d]
           2
      Sm 5[(2a 1 - d) + 5nd]
         =
      Sn   [(2a1 - d) + nd]
            A


            Sm
      for      to be independent of n
            Sn
      \ 2a1 – d = 0      Þ d = 2a1 Þ d = 6 Þ a2 = 9
      If d = 0    Þ a2 = a1 = 3
64.   Consider the parabola y2 = 8x. Let D1 be the area of the triangle formed by the end points of its
                                   æ1 ö
      latus rectum and the point P ç , 2 ÷ on the parabola, and D2 be the area of the triangle formed by
                                   è2 ø
                                                                               D1
     drawing tangents at P and at the end points of the latus rectum. Then D is
                                                                                 2

Sol. Ans. 2                                                         (2,4)
                                                                    A
     D1 = area of D PAA'                                (1/2,2)           y = 8x                                  2


                                                                                                P            Þa=2
                1 3
            =    .8. = 6
                2 2
                1
      D2 =        (D1)
                2                                                                                        A'
                                                                                                       (2,–4)
      (Using property : Area of triangle formed by tangents is always half of original triangle)
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           D1
      Þ       =2
          D2
65.   The positive integer value of n > 3 satisfying the equation
          1        1       1
              =          +        is
          æpö     æ 2p ö   æ 3p ö
      sin ç ÷ sin ç ÷ sin ç ÷
          ènø     è n ø    è n ø
Sol. Ans. 7
        1              1        1                           1              1        1
                =          +                     Þ                  -          =
            p           2p       3p                             p           3p       2p
      sin           sin      sin                         sin            sin      sin
            n            n        n                             n            n        n
           3p        p                2p     p
              - sin
            sin                  2cos sin
            n        n = 1             n     n = 1
      Þ                        Þ



      Þ 2cos


      Þ sin
            p
        sin sin
            n
               EN 3p
                   n



                  4p
                    2p
                    n
                        sin


                       sin

                     = sin
                           n
                            2p
                            n
                           2p

                           3p
                                  sin sin


                              = sin

                              Þ
                                4p
                                     p
                                     n
                                    3p
                                     n
                                          3p
                                           n
                                                sin




                                   = Kp + ( -1) K
                                                  3p
                                                    2p
                                                    n




                  n         n   n                  n
             LL
                                   p                                      1                   1 1 1
      If K = 2m Þ                    = 2mp               Þ          n=          Þ n=           , , ........
                                   n                                     2m                   2 4 6
                              7p                                                          7                             7 7
      If K = 2m + 1              = (2m + 1) p
                                  Þ                                      Þ n=                        Þ n = 7, , .......
                               n                                                        2m + 1                          3 5
            A

      Possible value of n is 7
                    æ -1 æ sin q ö ö            p       p                        d
66.  Let ƒ(q) = sin ç tan ç       ÷ ÷ , where - 4 < q < 4 . Then the value of          (ƒ(q)) is
                    è     è cos2q ø ø                                         d(tan q)
Sol. Ans. 1
                                     -1 æ sin q ö
      Let ƒ( q) = sin a where a = tan ç          ÷
                                        è cos 2q ø
                           sin q
      Þ      tan a =
                           cos 2q
                        sin q                     æ      æ p p öö
      Þ      sin a =          = tan q             çQ q Î ç - , ÷ ÷
                        cos q                     è      è 4 4 øø
      Þ      ƒ( q) = tan q
             d(ƒ(q))
      Þ               =1
             d(tan q)

67.   If z is any complex number satisfying |z – 3 – 2i| < 2, then the minimum value of |2z – 6 + 5i|
      is

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Sol. Ans. 5
                                                                                                                    Im
                                             æ    5 ö
      We have to find minimum value of 2 z - ç 3 - i ÷
                                             è    2 ø
                                                                                                                                          |z–3–2i| < 2
                                                      æ    5ö                        (3,2)
         = 2 × (minimum distance between z and point ç 3, - ÷ )
                                                      è    2ø
         = 2 × (distance between (3,0) and æ 3, - ö )
                                                 5                                                                                             Re
                                                                                  (3,0)
                                           ç       ÷
                                           è     2ø
                                                                                   (3,–5/2)
                 5
         =2×        = 5 units.
                 2
68. The minimum value of the sum of real numbers a–5, a–4, 3a–3, 1, a8 and a10 with a > 0 is
Sol. Ans. 8
     As a > 0
     and all the given terms are positive




      Þ
               EN
     hence considering A.M. > G.M. for given numbers :
      a -5 + a -4 + a -3 + a -3 + a -3 + a 8 + a10
                           7
                                                   ³ ( a .a .a .a .a .a .a )

          a -5 + a -4 + a -3 + a -3 + a -3 + a 8 + a10
                               7
                                                        -5 -4 -3 -3 -3 8 10 7




                                                       ³1                Þ
                                                                             1




                                                                                 (a   -5
                                                                                           + a -4 + 3a -3 + a 8 + a10 )
                                                                                                                             min
                                                                                                                                   =7
      where a–5 = a–4 = a–3 = a8 = a10 i.e. a = 1
             LL
      Þ (a–5 + a–4 + 3a–3 + a8 + a10 + 1)min = 8                          when a = 1
                                                                                                                         x

69.   Let ƒ : [1,¥) ® [2,¥) be a differentiable function such that f(1) = 2. If 6 ò ƒ(t)dt = 3x ƒ(x) - x 3
                                                                                                                         1

     for all x > 1, then the value of ƒ(2) is
            A

Sol. Ans. 6
     (Comment : The given relation does not hold for x =1, therefore it is not an identity. Hence there
     is an error in given question. The correct identity must be-)
        x
      6 ò f(t)dt = 3xf(x) – x 3 – 5, " x ³ 1
        1

      Now applying Newton Leibnitz theorem
      6ƒ(x) = 3xƒ'(x) – 3x2 + 3ƒ(x)
      Þ 3ƒ(x) = 3xƒ'(x) – 3x2
      Let y = ƒ(x)
                dy                    xdy - ydx                               æ yö
      Þ     x
                dx
                   - y = x2    Þ
                                         x2
                                                = dx              Þ       ò d ç x ÷ = ò dx
                                                                              è ø
          y
      Þ     = x+C    (where C is constant)
          x
      Þ y = x2 + Cx
      \ ƒ(x) = x2 + Cx
      Given ƒ(1) = 2 Þ C = 1
      \ ƒ(2) = 22 + 2 = 6
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